Problem: $f(x) = -6x^{3}-2x^{2}+6x+6+3(h(x))$ $h(n) = 4n^{3}-5n^{2}+4n$ $ f(h(1)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = 4(1^{3})-5(1^{2})+(4)(1)$ $h(1) = 3$ Now we know that $h(1) = 3$ . Let's solve for $f(h(1))$ , which is $f(3)$ $f(3) = -6(3^{3})-2(3^{2})+(6)(3)+6+3(h(3))$ To solve for the value of $f$ , we need to solve for the value of $h(3)$ $h(3) = 4(3^{3})-5(3^{2})+(4)(3)$ $h(3) = 75$ That means $f(3) = -6(3^{3})-2(3^{2})+(6)(3)+6+(3)(75)$ $f(3) = 69$